The work functions of two metals are 2.75 eV and 2eV respectively. If these are irradiated by photons of energy 3 e V, the ratio of maximum momenta of the photoelectrons emitted respectively by them is

Options

A. A)`1:2`
B. B)`1:3`
C. C)`1:4`
D. D)`2:1`

Answer

A

Explanation

Maximum kinetic energy for first metal
`KE_1 = 3 -2.75 = 0.25 eV`
Maximum kinetic energy for second metal
`KE_2 = 3 -2 = 1 eV`
The ratio between maximum momentum
`((p_1)/(p_2))_("max") = sqrt((KE_1)/(KE_2) ) " "(because P = sqrt(2 m KE))`
`=sqrt((0.25)/(1.00))`
`((p_1)/(p_2))_("max") = (1)/(2)`